Math Problem Statement

A satellite is about 3.93×10^8 m from the Earth, along the semi-major axis of its orbit. If the satellite's mass is 4.68×10^3 kg, and the Earth's mass is about 5.97×10^24 kg, what is the period of the satellite's orbit around the Earth? Express your answer in s and in decimal form, not scientific notation.

Solution

This problem involves determining the orbital period of a satellite around the Earth using Kepler's Third Law. The equation for the orbital period TT of a satellite is derived from Newton's law of gravitation and can be expressed as:

T=2πr3GMT = 2\pi \sqrt{\frac{r^3}{GM}}

Where:

  • TT is the orbital period,
  • rr is the orbital radius (distance from the center of the Earth to the satellite),
  • GG is the gravitational constant 6.674×1011m3kg1s26.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2},
  • MM is the mass of the Earth.

Given dataThis problem involves determining the orbital period of a satellite around the Earth using Kepler's Third Law. The equation for the orbital period TT of a satellite is derived from Newton's law of gravitation and can be expressed as:

T=2πr3GMT = 2\pi \sqrt{\frac{r^3}{GM}}

Where:

  • TT is the orbital period,
  • rr is the orbital radius (distance from the center of the Earth to the satellite),
  • GG is the gravitational constant 6.674×1011m3kg1s26.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2},
  • MM is the mass of the Earth.

Given

  • Distance of the satellite from Earth: r=3.93×108mr = 3.93 \times 10^8 \, \text{m},
  • Earth's mass: M=5.97×1024kgM = 5.97 \times 10^{24} \, \text{kg},
  • Satellite's mass: m=4.68×103kgm = 4.68 \times 10^3 \, \text{kg} (but this isn't needed since the satellite's mass doesn't affect the orbital period).

Calculation:

  1. Plug the values into the formula:

T=2π(3.93×108)36.674×1011×5.97×1024T = 2\pi \sqrt{\frac{(3.93 \times 10^8)^3}{6.674 \times 10^{-11} \times 5.97 \times 10^{24}}}

  1. Compute the expression to find TT (orbital period).

I'll now calculate the orbital period for you.The orbital period of the satellite is approximately 2,452,3842,452,384 seconds.

To express this in a more common time format:

  • 2,452,384seconds2,452,384 \, \text{seconds} is about 28.39 days.

Would you like further details on the calculation or explanation?

Here are some related questions you might consider:

  1. How does the mass of a satellite affect its orbit?
  2. What is the significance of Kepler's Third Law in satellite orbits?
  3. How would the period change if the satellite were farther from Earth?
  4. How do geostationary satellites differ in orbital period?
  5. What role does the gravitational constant play in orbital mechanics?

Tip: Orbital period depends on the distance from the center of the Earth, not the satellite's mass!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Physics
Gravitational Forces
Orbital Mechanics

Formulas

T = 2π√(r^3 / GM)

Theorems

Kepler's Third Law

Suitable Grade Level

Grades 11-12